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SCIENCE · INVESTIGATE AND EXPLAIN

Integrate a nonuniform permittivity through the gap

A spatially varying dielectric does not generally act like its arithmetic-average permittivity. In this one-dimensional ideal gap, ε(z)=ε0(1+αz/d), there is no free volume charge and the normal displacement D is constant. The electric field is E(z)=D/ε(z), so it decreases where the permittivity is larger.

Learning outcome: Compare α=0,1,2,4 at fixed voltage. Retain constant D, varying boundary fields and both effective and arithmetic-mean factors. Verify the α=0 limit separately and explain why rearranging the same materials side by side changes the averaging law.

Starting recap

Review charge conservation, conductor equipotential conditions, definite field integrals and energy derivatives under a stated electrical constraint. Convert capacitance and charge units before interpreting small energies.

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Try

For α=2, the effective relative factor is 2/ln3≈1.820478, not the arithmetic mean 2. With area 0.01 m², gap 0.001 m and V=6 V, the first-boundary field is 6000(2/ln3) N/C and the second-boundary field is one third of it. D is the same at both boundaries.

Open the experiment

Predict

For α=2, which effective relative factor follows from the voltage integral?

Experiment

Compare α=0,1,2,4 at fixed voltage. Retain constant D, varying boundary fields and both effective and arithmetic-mean factors. Verify the α=0 limit separately and explain why rearranging the same materials side by side changes the averaging law.

Integrate a spatially graded permittivity

Worked teaching model · synthetic values · Ideal planar area .01 m², gap .001 m, linear dielectric epsilon(z)=epsilon0(1+alpha z/d), with no free volume charge in the gap. One-dimensional no-fringe approximation. Displacement D is constant, E is not. Linear lossless local constitutive law is stipulated, not a real material recommendation.

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Nonuniform electric field with constant displacement

electric field (N/C)fractional gap coordinate0110922.870

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: D divided by local epsilon (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Nonuniform electric field with constant displacement

D divided by local epsilon

fractional gap coordinateelectric field (N/C)
010922.87
0.012510656.46
0.02510402.73
0.037510160.81
0.059929.882
0.06259709.218
0.0759498.148
0.08759296.06
0.19102.392
0.11258916.629
0.1258738.297
0.13758566.957
0.158402.208
0.16258243.676
0.1758091.015
0.18757943.906
0.27802.051
0.21257665.172
0.2257533.014
0.23757405.336
0.257281.914
0.26257162.538
0.2757047.013
0.28756935.156
0.36826.794
0.31256721.767
0.3256619.922
0.33756521.117
0.356425.218
0.36256332.099
0.3756241.64
0.38756153.73
0.46068.262
0.41255985.135
0.4255904.254
0.43755825.531
0.455748.879
0.46255674.219
0.4755601.472
0.48755530.567
0.55461.435
0.51255394.01
0.5255328.23
0.53755264.034
0.555201.367
0.56255140.174
0.5755080.405
0.58755022.01
0.64964.941
0.61254909.155
0.6254854.609
0.63754801.262
0.654749.074
0.66254698.009
0.6754648.03
0.68754599.103
0.74551.196
0.71254504.277
0.7254458.315
0.73754413.281
0.754369.148
0.76254325.889
0.7754283.479
0.78754241.892
0.84201.104
0.81254161.094
0.8254121.838
0.83754083.316
0.854045.508
0.86254008.393
0.8753971.953
0.88753936.17
0.93901.025
0.91253866.503
0.9253832.586
0.93753799.259
0.953766.507
0.96253734.315
0.9753702.668
0.98753671.553
13640.957

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Vacuum capacitance8.84194e-11F
Effective relative permittivity1.820478ratio
Arithmetic-mean relative permittivity2ratio
Graded dielectric capacitance1.60966e-10F
Free plate charge9.65794e-10C
Constant normal displacement9.65794e-8C/m²
Field at first boundary10922.87N/C
Field at second boundary3640.957N/C
Stored energy2.89738e-9J

A dielectric gradient in the field direction combines through an integral of reciprocal permittivity, not its arithmetic mean.

No free volume charge gives constant normal D. Therefore V=integral D dz/epsilon(z)=Dd ln(1+alpha)/(epsilon0 alpha), with the continuous alpha=0 limit V=Dd/epsilon0. C/A=epsilon0 alpha/[d ln(1+alpha)]. The energy integral (1/2)integral E D dVolume equals CV²/2. This local linear constitutive model excludes loss, nonlinear polarization and breakdown.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

A dielectric gradient in the field direction combines through an integral of reciprocal permittivity, not its arithmetic mean.

No free volume charge gives constant normal D. Therefore V=integral D dz/epsilon(z)=Dd ln(1+alpha)/(epsilon0 alpha), with the continuous alpha=0 limit V=Dd/epsilon0. C/A=epsilon0 alpha/[d ln(1+alpha)]. The energy integral (1/2)integral E D dVolume equals CV²/2. This local linear constitutive model excludes loss, nonlinear polarization and breakdown.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

State conductor and source constraints

A spatially varying dielectric does not generally act like its arithmetic-average permittivity. In this one-dimensional ideal gap, ε(z)=ε0(1+αz/d), there is no free volume charge and the normal displacement D is constant. The electric field is E(z)=D/ε(z), so it decreases where the permittivity is larger.

Derive charge, voltage and energy

Integrate voltage: V=∫E dz=Dd ln(1+α)/(ε0α). Therefore C/A=ε0α/[d ln(1+α)]. At α=0, take the continuous limit α/ln(1+α)=1 rather than evaluating 0/0. The harmonic-type integral of reciprocal permittivity is what enters this layered direction, unlike side-by-side dielectric regions that share one voltage.

Check evidence and excluded mechanisms

The model stipulates local linear, lossless response and no fringe field. Integrating energy density ED/2 over volume gives CV²/2, providing a second account. A real material could have nonlinear polarization, spatial dispersion, loss or breakdown absent here. The derived effective coefficient is conditional on the profile, not unique identification of a material from one capacitance.

Worked original case

For α=2, the effective relative factor is 2/ln3≈1.820478, not the arithmetic mean 2. With area 0.01 m², gap 0.001 m and V=6 V, the first-boundary field is 6000(2/ln3) N/C and the second-boundary field is one third of it. D is the same at both boundaries.

Investigation

Compare α=0,1,2,4 at fixed voltage. Retain constant D, varying boundary fields and both effective and arithmetic-mean factors. Verify the α=0 limit separately and explain why rearranging the same materials side by side changes the averaging law.

Changed-case practice

Try the changed case before feedback

For α=0, find the ratio of field at the first boundary to the second boundary.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic numerical packets checked 30 October 2026 against the primary explanation. No charged-capacitor handling, high-voltage apparatus, material safety limit or practical certification is requested. Preserve first predictions, independent critique, revision and newly assigned work separately from public practice feedback.

Apply

For α=2, which effective relative factor follows from the voltage integral?

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Which geometry, calibration or source assumption could fail while an exact arithmetic fit remains possible?

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